{"id":186223,"date":"2025-01-24T14:11:17","date_gmt":"2025-01-24T14:11:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186223"},"modified":"2025-01-24T14:11:19","modified_gmt":"2025-01-24T14:11:19","slug":"annotate-the-following-c-and-h-nmr-and-label-the-peaks-of-the-following-compounds-and-provide-the-structure","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/24\/annotate-the-following-c-and-h-nmr-and-label-the-peaks-of-the-following-compounds-and-provide-the-structure\/","title":{"rendered":"Annotate the following C and H-NMR and label the peaks of the following compounds and provide the structure"},"content":{"rendered":"\n<p>Annotate the following C and H-NMR and label the peaks of the following compounds and provide the structure:<\/p>\n\n\n\n<p>2-methoxy-6-(p-tolyliminomethyl)-phenol:<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To annotate the C-NMR and H-NMR spectra of 2-methoxy-6-(p-tolyliminomethyl)-phenol, we first need to understand its structure. This compound consists of a phenol ring substituted with a methoxy group (\u2013OCH\u2083) at position 2, an iminomethyl group (\u2013CH\u2082=N\u2013) at position 6, and a p-tolyl group (\u2013C\u2086H\u2084-CH\u2083) attached to the imino group.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Structure of 2-methoxy-6-(p-tolyliminomethyl)-phenol:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Phenol ring<\/strong>: A benzene ring with a hydroxyl group (-OH) at position 1.<\/li>\n\n\n\n<li><strong>Methoxy group<\/strong> (-OCH\u2083) at position 2.<\/li>\n\n\n\n<li><strong>Iminomethyl group<\/strong> (-CH\u2082=N-) at position 6 of the phenol ring.<\/li>\n\n\n\n<li><strong>p-Tolyl group<\/strong> (-C\u2086H\u2084-CH\u2083) attached to the iminomethyl group.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">C-NMR Spectrum:<\/h3>\n\n\n\n<p>The C-NMR spectrum provides the chemical environments of the carbon atoms in the compound.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Aromatic region<\/strong> (around 110\u2013160 ppm):<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C-2<\/strong>: Methoxy group (-OCH\u2083) attached to C-2: Typically, this will appear around 55-60 ppm for the methoxy group carbon (\u2013OCH\u2083).<\/li>\n\n\n\n<li><strong>C-6<\/strong>: Attached to the iminomethyl group: This carbon resonates around 130\u2013140 ppm (due to the electron-withdrawing effect of the imino group).<\/li>\n\n\n\n<li><strong>Other aromatic carbons<\/strong>: The remaining carbons in the aromatic ring (C-3, C-4, C-5, C-7, C-8) will fall within the 110\u2013140 ppm range.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Aliphatic region<\/strong> (around 40\u201360 ppm):<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Iminomethyl group (-CH\u2082=N)<\/strong>: The carbon attached to the nitrogen in the iminomethyl group will appear around 40\u201350 ppm.<\/li>\n\n\n\n<li><strong>p-Tolyl methyl group (\u2013CH\u2083)<\/strong>: The methyl group of the p-tolyl group will appear around 20\u201325 ppm.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">H-NMR Spectrum:<\/h3>\n\n\n\n<p>The H-NMR spectrum provides the chemical shifts corresponding to protons in the compound.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Aromatic protons<\/strong> (around 6.5\u20138 ppm):<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Protons on the aromatic ring, especially on positions 3, 4, and 5, will resonate around 6.5\u20138 ppm.<\/li>\n\n\n\n<li>The pattern of these peaks will likely be multiplets due to the different couplings of the protons.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Methoxy proton<\/strong> (around 3.5\u20134.0 ppm):<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The proton of the methoxy group (-OCH\u2083) typically appears around 3.5\u20134.0 ppm.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Iminomethyl group<\/strong> (around 4.5\u20135.5 ppm):<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The \u2013CH\u2082 group in the iminomethyl will appear as a doublet or triplet around 4.5\u20135.5 ppm, depending on the coupling constants.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>p-Tolyl methyl protons<\/strong> (around 2.2\u20132.5 ppm):<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The methyl protons of the p-tolyl group will resonate around 2.2\u20132.5 ppm, showing a doublet due to the coupling with the adjacent aromatic protons.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<p>The structure of 2-methoxy-6-(p-tolyliminomethyl)-phenol consists of a methoxy group, an iminomethyl group, and a p-tolyl group, which influence the chemical shifts in both the C-NMR and H-NMR spectra. The annotation of the peaks will involve identifying the shifts corresponding to the aromatic, methoxy, iminomethyl, and p-tolyl groups. These spectra help to confirm the structure based on the expected chemical environments of the atoms in the molecule.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Annotate the following C and H-NMR and label the peaks of the following compounds and provide the structure: 2-methoxy-6-(p-tolyliminomethyl)-phenol: The Correct Answer and Explanation is : To annotate the C-NMR and H-NMR spectra of 2-methoxy-6-(p-tolyliminomethyl)-phenol, we first need to understand its structure. This compound consists of a phenol ring substituted with a methoxy group (\u2013OCH\u2083) [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-186223","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186223","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=186223"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186223\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186223"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186223"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186223"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}